The question and answer that I have problem understanding is as follows:

A helicopter of mass 800 kg rises to a height of 170 metres in 20 seconds, before setting off in horizontal flight. Calculate the potential energy gain of the helicopter, and hence estimate the mean power of its engine. State a form of kinetic energy that has been ignored in this model.

The first and second questions are easy to me.

Provided that g=10,

1) W2h2 - W1h1 = 8000 X 170 - 8000 X 0 = 1,360,000J

2) 1,360,000/20 = 68kW

But what I am struggling with is the third question - State a form of kinetic energy that has been ignored in this model.

The answer provided is "The kinetic energy of the rotor", and I just don't quite understand why it can be the answer for this question.

In this model, you need 68kW of power to lift up the helicopter 170 m in 20 seconds, right?

And the engine provides that amount of power "through" its rotor, correct?

So the rotor requires the same amount of power (i.e. 68kW) to lift up the helicopter 170 m in 20 seconds, correct?

Then why do we have to calculate the kinetic engergy of the rotor, which is the energy to rotate the rotor, and add it to the potential energy to figure out the total power required to lift up the helicopter???

Maybe there is something wrong with my understading of potential energy and kinetic energy and how they relate to power..

Anyway, I would much appreciate it if somone out there can help me to undersand this answer to this question in a way that it is easy to follow even for a beginner of physics like me.

. Calculate the potential energy gain of the helicopter,....... State a form of kinetic energy that has been ignored in this model. [/I]

The question seems to want you to indicate where there is a change in kinetic energy from when the helicopter is on the ground to when it is 170 m up...

Well ... where is there kinetic energy ??? there is some in the rotor ...

But it's not at all guaranteed that the speed of the rotor has changed .

The rotor can be spinning very fast on the ground , but may generate no lift until the pitch of the blades is changed, when this is done we have lift , the blades will tend to slow , but the engine may deliver more power to keep them fast....

So we do not know if the rotor is spinning faster of slower at 170m. or if it's changed at all.

Thank you for your explanation with illustration, which really helped me to understand what was missing in my attempt to make sense of the answer provided in the textbook.

I did not even know that there is such a thing as "tail rotor" on helicopter!

Yes, you are right. The engine requires the power to keep the helicopter's tail rotor moving on top of the power to lift up its body so that it will not keep going around and around rotating up in the air!